Mehr62
Answered

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If the distance between two positive point charges is tripled, then the strength of the electrostatic repulsion between them will decrease by a factor of of how much?

Sagot :

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Coulomb's Law:
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
k is a constant
q1 and q2 are charges
r is the distance


[tex]F_1=\frac{kq_1q_2}{r^2}[/tex]

the distance is tripled:
[tex]F_2=\frac{kq_1q_2}{(3r)^2}=\frac{kq_1q_2}{9r^2}[/tex]


[tex]\frac{\frac{kq_1q_2}{9r^2}}{\frac{kq_1q_2}{r^2}}= \frac{r^2}{9r^2}=\frac{1}{9}[/tex]

The strength of the electrostatic repulsion will decrease by a factor of 9.
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